all norms are equivalent on a fdvs over a complete field

Stream: maths

[Kevin Buzzard]

It's well-known (e.g. Cassels-Froehlich lemma p52 section 8 of Cassels' article) that if k is a complete normed field and V is a finite-dimensional vector space over k then all k-norms on V are equivalent, and hence induce the same topology. Do we have this in mathlib? Here are some naive questions about generalizations of this.

1) Is there some variant of this for R a complete normed ring and V say a finite free R-module?
2) Is this a purely topological statement? If k is a complete topological field then is there only one topology on a finite-dimensional k-vector space V making addition and scalar multiplication continuous?
3) Same as 2 but for rings.

I have no real feeling for how optimistic I'm being here. Looking at Cassels' proof and trying to push it through for topologies on complete topological rings, the easy direction goes like this (and doesn't assume completeness).

R a complete top ring, V a finite free R-module (let's stick with this case for now). If we choose a basis for V then V=R^n and we can give it the product topology; this is the R-module topology mV for V by results we have in mathlib. Now put an arbitrary topology tV on V making it into a topological ring. We know the identity function from mV to tV is continuous i.e. mV<=tV, because mV is defined to be the inf of the R-module topologies on V.

So what's left is the work, namely proving (if it's true) that tV<=mV. One does have to use completeness somehow: the standard counterexample is that if you give $\mathbb{Q}$ and $\mathbb{Q}(\sqrt{2})$ the topologies coming from $\R$ (choose the positive square root of 2) then the map $\mathbb{Q}^2\to\mathbb{Q}(\sqrt{2})$ sending $(a,b)$ to $a+b\sqrt{2}$ is continuous but doesn't have a continuous inverse. Does completeness somehow save us in the topological case, or are these really theorems about topologies coming from norms?

[Yakov Pechersky]

Doesn't help with generalization, but KConrad also has notes on this https://kconrad.math.uconn.edu/blurbs/gradnumthy/equivnorms.pdf

[Andrew Yang]

If there are no conditions on V except continuity of addition and scalar multiplication then the indiscrete topology probably also satisfies this?

[Yakov Pechersky]

Then it's not normed because norm eq 0 doesn't imply 0?

[Andrew Yang]

I’m referring to 2 where Kevin asked if it is a topological phenomenon.

[Yakov Pechersky]

Ah. Well if we're talking about completions then probably the uniform space structure is relevant. And metrics would be equivalent only if the uniform structures are equivalent. But one could have equivalent topologies without equivalent uniformities. https://math.stackexchange.com/a/793828

[Oliver Nash]

Regarding "Do we have this in mathlib?" and point 2, see docs#FiniteDimensional.nonempty_continuousLinearEquiv_iff_finrank_eq or docs#LinearMap.continuous_of_finiteDimensional showing that the answers are "yes" and "yes, provided you assume Hausdorff".

Oh you don't have a norm on the field, sorry should have read more carefully.

[Kevin Buzzard]

Andrew Yang said:

If there are no conditions on V except continuity of addition and scalar multiplication then the indiscrete topology probably also satisfies this?

Darn it, I think you're right :-/ Discrete topology doesn't work because scalar multiplication isn't continuous in general, but indiscrete topology looks fine. So what other topological property is implied by normed field and normed vector space? I guess one topological property which rules out the indiscrete topology is that for all v in V the induced map R -> V sending r to r.v is strict, that is, the quotient topology on the image (coming from R) equals the subspace topology (coming from V). Is that enough to save us?

Revised version of most optimistic question:

Q) If $R$ is a topological ring, complete as a uniform additive group, and if $V$ is a finite free (or more generally finite projective, or finite flat, but let's stick to free for now) $R$-module with the property that for all $v\in V$ the subspace and quotient topologies on $Rv$ coincide, then does $V$ have the product (or more generally the R-module) topology?

[Kevin Buzzard]

Re Oliver's message: wow, I am amazed at the role Hausdorff is playing here. As you pointed out in an earlier version of that message, the key point is docs#unique_topology_of_t2. Here's how I think about this kind of thing now. Let $R$ be any fixed topological ring, and let's define $M$ to be $R$ without its topology. Then we have addition $M\times M\to M$ and multiplication $R\times M\to M$ and we can ask what possible topologies on $M$ there are, which make both of these maps continuous (putting the product topology on the source). To put it another way, can we understand the topologies on $M$ making $M$ into a topological $R$-module? The fact that $R$ is a topological ring means that one example of such a topology on $M$ is the one coming from $M=R$. It can be checked docs#TopologicalSemiring.toIsModuleTopology that in fact this is the inf of all the possible topologies on $M$ making $M$ into a topological $R$-module, but there are others (e.g. the indiscrete topology). Note that it's not true that if topologies $t_1$ and $t_2$ work then any topology $t$ with $t_1\leq t\leq t_2$ works; it's more subtle than that. So in this generality we've seen two topologies which work and it would not surprise me if the pullback of the quotient topology on $R/I$ would also work for any two-sided ideal works, although I didn't check this. Of course none of these are in general Hausdorff, as the closure of {0} would be $I$. The extraordinary (in my mind) result flagged by Oliver is that if $R$ is a nontrivially normed field then the only Hausdorff topology on $M$ is just $R$'s topology; I was expecting there to be a whole zoo of possibilities in general.

[Mitchell Lee]

Let $R$ be a topological ring. If $M = R^m$ is a finite free $R$-module and all linear maps $R \to M$ and $M \to R$ are continous, then $M$ must have the module topology.

In fact it suffices for all the projections $p_j\colon M \to R$ and inclusions $i_j \colon R \to M$ to be continuous. Then $p_j i_{j'}$ is the identity if $j = j'$ and $0$ otherwise, and moreover $i_1 p_1 + \cdots + i_m p_m$ is the identity on $M$. This characterizes $M$ as the direct sum of $m$ copies of $R$ in the additive category of topological $R$-modules, so the topology is uniquely determined.

For finite dimensional normed vector spaces over a field, every linear map is continuous, so this implies the result you were asking about.

[Mitchell Lee]

I am not sure about (Q) though.

[Andrew Yang]

I think an analogous result holds for compact Hausdorff local rings that there exists a unique compact Hausdorff topology on R that makes it a topological module over itself.

[Sébastien Gouëzel]

I don't know if this has been said clearly earlier, so let me say it again: over a complete nontrivially normed field, a linear map from a finite-dimensional Hausdorff topological vector space E is always continuous (docs#LinearMap.continuous_of_finiteDimensional). This implies that two topologies making E into a Hausdorff topological vector space have to coincide (see e.g. docs#LinearEquiv.toContinuousLinearEquiv).

I have no idea what happens outside the context of normed fields.